IODS course project
Md Karim Uddin
Introduction to Open Data Science - Course Project
About the project
This Course has opened up a new dimension for learning data science for the persons with different background
# This is a so-called "R chunk" where you can write R code.
date()## [1] "Thu Nov 19 17:49:08 2020"I came to know about this course in weboodi
Regression and model validation
##by Md Karim Uddin
Part1
date()## [1] "Thu Nov 19 17:49:08 2020"Reading the dataset to start the analysis.
learning2014<-read.table( "learning2014.csv", header = TRUE,sep = ",", stringsAsFactors = FALSE)dim(learning2014) ## [1] 166 7We can see that the table consists of 166 observations and 7 variables (i.e. 166 rows and 7 columns when viewed as a table).
Now, let’s look at the structure of the dataset
str(learning2014)## 'data.frame': 166 obs. of 7 variables:
## $ gender : chr "F" "M" "F" "M" ...
## $ age : int 53 55 49 53 49 38 50 37 37 42 ...
## $ attitude: num 3.7 3.1 2.5 3.5 3.7 3.8 3.5 2.9 3.8 2.1 ...
## $ deep : num 3.58 2.92 3.5 3.5 3.67 ...
## $ stra : num 3.38 2.75 3.62 3.12 3.62 ...
## $ surf : num 2.58 3.17 2.25 2.25 2.83 ...
## $ points : int 25 12 24 10 22 21 21 31 24 26 ...As we can see, the data set consists of 166 observations of 7 variables. The column “gender” gives the respondents gender (M = male, F= Female) and the column “Age” their age in years. The column “attitude” describes the respondents attitude towards statistics and the column “points” their totalt points in an exam. The columns “deep”, “stra” and “sur” give the mean points to questions about deep, strategic and surface learning, respectively.
Let’s check the summary of the table, to know how the data is distributed.
summary(learning2014)## gender age attitude deep
## Length:166 Min. :17.00 Min. :1.400 Min. :1.583
## Class :character 1st Qu.:21.00 1st Qu.:2.600 1st Qu.:3.333
## Mode :character Median :22.00 Median :3.200 Median :3.667
## Mean :25.51 Mean :3.143 Mean :3.680
## 3rd Qu.:27.00 3rd Qu.:3.700 3rd Qu.:4.083
## Max. :55.00 Max. :5.000 Max. :4.917
## stra surf points
## Min. :1.250 Min. :1.583 Min. : 7.00
## 1st Qu.:2.625 1st Qu.:2.417 1st Qu.:19.00
## Median :3.188 Median :2.833 Median :23.00
## Mean :3.121 Mean :2.787 Mean :22.72
## 3rd Qu.:3.625 3rd Qu.:3.167 3rd Qu.:27.75
## Max. :5.000 Max. :4.333 Max. :33.00female <- sum(learning2014$gender == "F")
male <- sum(learning2014$gender == "M")
print(paste("The datset contains", toString(male), "male and", toString(female), "female students"))## [1] "The datset contains 56 male and 110 female students"From above we can see the means, medians, ranges (min, max) and quartiles of the variables. For example, the students’ ages range from 21 to 55 with a mean of 25.5 and a median of 22.
110 of the students are female and 56 are male
library(GGally)## Loading required package: ggplot2## Registered S3 method overwritten by 'GGally':
## method from
## +.gg ggplot2library(ggplot2)p1 <- ggpairs(learning2014, mapping = aes(), lower = list(combo = wrap("facethist", bins = 20)))p1
Based on this, points seem to be positively correlated with attitude (correlation coefficient 0.437, then strategic learning (0.146) and finally negatively correlated with surface learning (-0.144).
Lets create a linear model including all these three variables and check the model.
my_model <- lm(points ~ attitude + stra + surf, data = learning2014)
summary(my_model)##
## Call:
## lm(formula = points ~ attitude + stra + surf, data = learning2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17.1550 -3.4346 0.5156 3.6401 10.8952
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.0171 3.6837 2.991 0.00322 **
## attitude 3.3952 0.5741 5.913 1.93e-08 ***
## stra 0.8531 0.5416 1.575 0.11716
## surf -0.5861 0.8014 -0.731 0.46563
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.296 on 162 degrees of freedom
## Multiple R-squared: 0.2074, Adjusted R-squared: 0.1927
## F-statistic: 14.13 on 3 and 162 DF, p-value: 3.156e-08We can see that attitude is the only significant variable in the model (i.e p < 0.05). The coefficient for attitude is 3.4 (i.e. when an increase of one in attitude score predicts an increase of 3.4 in exam points). Let’s get rid of the non-significant variables and fit a new model.
new_model <- lm(points ~ attitude, data = learning2014)
summary(new_model)##
## Call:
## lm(formula = points ~ attitude, data = learning2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -16.9763 -3.2119 0.4339 4.1534 10.6645
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.6372 1.8303 6.358 1.95e-09 ***
## attitude 3.5255 0.5674 6.214 4.12e-09 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.32 on 164 degrees of freedom
## Multiple R-squared: 0.1906, Adjusted R-squared: 0.1856
## F-statistic: 38.61 on 1 and 164 DF, p-value: 4.119e-09Now the model has a R^2 of 0.1906 and is still signficant (p<0.05). The R^2 value means that it is able to explain 19.06 % of the variance in the points.
Next we’ll check the diagnostics
plot(new_model, which = c(1, 2, 5))
The first plots shows the residuals vs the fitted values, i.e. how far away each data point is from our predicted model. THe data point should be equally far from our predicted model throughout the model (homoscedasticity). The residuals are equally distributed througout the predicted values.
The second plot shows how residuals are distributed. The better they follow the Q-Q line, the more normal is their distributed. The residuals are sufficiently normal.
The third plot shows the residuals vs leverage. The further to the right a point is, the more leverage it has, i.e. affects the model. This is useful to detect significant outliers. THe model does not suffer form significant outlie
3.Logistic regression
##by Md Karim Uddin
Part1
date()## [1] "Thu Nov 19 17:49:15 2020"alc<-read.table( "alc.csv", header = TRUE,sep = ",", stringsAsFactors = FALSE)Dimension of the data
dim(alc) ## [1] 382 35It can bee seen that , the dataset has 382 observations and 35 variables.
str(alc) ## 'data.frame': 382 obs. of 35 variables:
## $ school : chr "GP" "GP" "GP" "GP" ...
## $ sex : chr "F" "F" "F" "F" ...
## $ age : int 18 17 15 15 16 16 16 17 15 15 ...
## $ address : chr "U" "U" "U" "U" ...
## $ famsize : chr "GT3" "GT3" "LE3" "GT3" ...
## $ Pstatus : chr "A" "T" "T" "T" ...
## $ Medu : int 4 1 1 4 3 4 2 4 3 3 ...
## $ Fedu : int 4 1 1 2 3 3 2 4 2 4 ...
## $ Mjob : chr "at_home" "at_home" "at_home" "health" ...
## $ Fjob : chr "teacher" "other" "other" "services" ...
## $ reason : chr "course" "course" "other" "home" ...
## $ nursery : chr "yes" "no" "yes" "yes" ...
## $ internet : chr "no" "yes" "yes" "yes" ...
## $ guardian : chr "mother" "father" "mother" "mother" ...
## $ traveltime: int 2 1 1 1 1 1 1 2 1 1 ...
## $ studytime : int 2 2 2 3 2 2 2 2 2 2 ...
## $ failures : int 0 0 2 0 0 0 0 0 0 0 ...
## $ schoolsup : chr "yes" "no" "yes" "no" ...
## $ famsup : chr "no" "yes" "no" "yes" ...
## $ paid : chr "no" "no" "yes" "yes" ...
## $ activities: chr "no" "no" "no" "yes" ...
## $ higher : chr "yes" "yes" "yes" "yes" ...
## $ romantic : chr "no" "no" "no" "yes" ...
## $ famrel : int 4 5 4 3 4 5 4 4 4 5 ...
## $ freetime : int 3 3 3 2 3 4 4 1 2 5 ...
## $ goout : int 4 3 2 2 2 2 4 4 2 1 ...
## $ Dalc : int 1 1 2 1 1 1 1 1 1 1 ...
## $ Walc : int 1 1 3 1 2 2 1 1 1 1 ...
## $ health : int 3 3 3 5 5 5 3 1 1 5 ...
## $ absences : int 5 3 8 1 2 8 0 4 0 0 ...
## $ G1 : int 2 7 10 14 8 14 12 8 16 13 ...
## $ G2 : int 8 8 10 14 12 14 12 9 17 14 ...
## $ G3 : int 8 8 11 14 12 14 12 10 18 14 ...
## $ alc_use : num 1 1 2.5 1 1.5 1.5 1 1 1 1 ...
## $ high_use : logi FALSE FALSE TRUE FALSE FALSE FALSE ...initialize a plot of high_use and G3
library(ggplot2)
g1 <- ggplot(alc, aes(x = high_use, y = G3, col = sex))
g1 + geom_boxplot() + ylab("grade")
g2 <- ggplot(alc, aes(x = high_use, y = absences, col = sex))
# define the plot as a boxplot and draw it
g2 + geom_boxplot() + ggtitle("Student absences by alcohol consumption and sex")
#logistic regression model
m <- glm(high_use ~ failures + absences + sex, data = alc, family = "binomial")
# print out a summary of the model
summary(m)##
## Call:
## glm(formula = high_use ~ failures + absences + sex, family = "binomial",
## data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.1855 -0.8371 -0.6000 1.1020 2.0209
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.90297 0.22626 -8.411 < 2e-16 ***
## failures 0.45082 0.18992 2.374 0.017611 *
## absences 0.09322 0.02295 4.063 4.85e-05 ***
## sexM 0.94117 0.24200 3.889 0.000101 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 424.40 on 378 degrees of freedom
## AIC: 432.4
##
## Number of Fisher Scoring iterations: 4library(knitr)
coef(m)## (Intercept) failures absences sexM
## -1.90296550 0.45081981 0.09321999 0.94116602library(magrittr)
library(knitr)m <- glm(high_use ~ failures + absences + sex, data = alc, family = "binomial") OR 2.5 % 97.5 %(Intercept) 0.1491257 0.09395441 0.228611 failures 1.5695984 1.08339644 2.294737 absences 1.0977032 1.05169654 1.150848 sexM 2.5629682 1.60381392 4.149405
Predictive power of the model-1
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionm <- glm(high_use ~ failures + absences + sex, data = alc, family = "binomial")
# predict() the probability of high_use
probabilities <- predict(m, type = "response")
# add the predicted probabilities to 'alc'
alc <- mutate(alc, probability = probabilities)
# use the probabilities to make a prediction of high_use
alc <- mutate(alc, prediction = probability > 0.5)
# see the last ten original classes, predicted probabilities, and class predictions
select(alc, failures, absences, sex, high_use, probability, prediction) %>% tail(10)## failures absences sex high_use probability prediction
## 373 1 0 M FALSE 0.3749639 FALSE
## 374 1 7 M TRUE 0.5353311 TRUE
## 375 0 1 F FALSE 0.1406689 FALSE
## 376 0 6 F FALSE 0.2069112 FALSE
## 377 1 2 F FALSE 0.2199932 FALSE
## 378 0 2 F FALSE 0.1523192 FALSE
## 379 2 2 F FALSE 0.3068503 FALSE
## 380 0 3 F FALSE 0.1647495 FALSE
## 381 0 4 M TRUE 0.3568828 FALSE
## 382 0 2 M TRUE 0.3153209 FALSE# tabulate the target variable versus the predictions
table(high_use = alc$high_use, prediction = alc$prediction)## prediction
## high_use FALSE TRUE
## FALSE 259 9
## TRUE 84 30Predictive power of the model-2
#initialize a plot of 'high_use' versus 'probability' in 'alc'
g <- ggplot(alc, aes(x = probability, y = high_use, col = prediction))
# define the geom as points and draw the plot
g + geom_point()
# tabulate the target variable versus the predictions
table(high_use = alc$high_use, prediction = alc$prediction) %>% prop.table %>% addmargins## prediction
## high_use FALSE TRUE Sum
## FALSE 0.67801047 0.02356021 0.70157068
## TRUE 0.21989529 0.07853403 0.29842932
## Sum 0.89790576 0.10209424 1.00000000Accuracy and loss functions
#the logistic regression model m and dataset alc with predictions are available
# define a loss function (average prediction error)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
# call loss_func to compute the average number of wrong predictions in the (training) data
loss_func(class = alc$high_use, prob = alc$probability)## [1] 0.2434555#Cross Validation
# the logistic regression model m and dataset alc (with predictions) are available
# define a loss function (average prediction error)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
# compute the average number of wrong predictions in the (training) data
loss_func(class = alc$high_use, prob = alc$probability)## [1] 0.243455510-fold cross-validation
library(boot)
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10)
# average number of wrong predictions in the cross validation
cv$delta[1]## [1] 0.26439794.Clustering and classification
by Md Karim Uddin
date()## [1] "Thu Nov 19 17:49:17 2020"Lets access the MASS library and “Boston” data
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:dplyr':
##
## selectdata("Boston")Now, we can explore the data set
# explore the dataset
str(Boston)## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00We can see that “Boston”data frame has 506 rows and 14 columns.The data contains information on per capita crime rate, average number of rooms per dwelling and so on.
Lets see the possible graphical presentation can be made from the existing variables
pairs(Boston)
Lets install corrplot package for graphical presentation
library(magrittr)
library(knitr)
library(plyr)## ------------------------------------------------------------------------------## You have loaded plyr after dplyr - this is likely to cause problems.
## If you need functions from both plyr and dplyr, please load plyr first, then dplyr:
## library(plyr); library(dplyr)## ------------------------------------------------------------------------------##
## Attaching package: 'plyr'## The following objects are masked from 'package:dplyr':
##
## arrange, count, desc, failwith, id, mutate, rename, summarise,
## summarizelibrary(corrplot)## corrplot 0.84 loadedLets access the tidyverse and MASS library
library(tidyverse)## -- Attaching packages --------------------------------------- tidyverse 1.3.0 --## v tibble 3.0.4 v purrr 0.3.4
## v tidyr 1.1.2 v stringr 1.4.0
## v readr 1.4.0 v forcats 0.5.0## -- Conflicts ------------------------------------------ tidyverse_conflicts() --
## x plyr::arrange() masks dplyr::arrange()
## x purrr::compact() masks plyr::compact()
## x plyr::count() masks dplyr::count()
## x tidyr::extract() masks magrittr::extract()
## x plyr::failwith() masks dplyr::failwith()
## x dplyr::filter() masks stats::filter()
## x plyr::id() masks dplyr::id()
## x dplyr::lag() masks stats::lag()
## x plyr::mutate() masks dplyr::mutate()
## x plyr::rename() masks dplyr::rename()
## x MASS::select() masks dplyr::select()
## x purrr::set_names() masks magrittr::set_names()
## x plyr::summarise() masks dplyr::summarise()
## x plyr::summarize() masks dplyr::summarize()library(MASS)Lets make the correlation matrix and visualize
cor_matrix<-cor(Boston)
cor_matrix## crim zn indus chas nox
## crim 1.00000000 -0.20046922 0.40658341 -0.055891582 0.42097171
## zn -0.20046922 1.00000000 -0.53382819 -0.042696719 -0.51660371
## indus 0.40658341 -0.53382819 1.00000000 0.062938027 0.76365145
## chas -0.05589158 -0.04269672 0.06293803 1.000000000 0.09120281
## nox 0.42097171 -0.51660371 0.76365145 0.091202807 1.00000000
## rm -0.21924670 0.31199059 -0.39167585 0.091251225 -0.30218819
## age 0.35273425 -0.56953734 0.64477851 0.086517774 0.73147010
## dis -0.37967009 0.66440822 -0.70802699 -0.099175780 -0.76923011
## rad 0.62550515 -0.31194783 0.59512927 -0.007368241 0.61144056
## tax 0.58276431 -0.31456332 0.72076018 -0.035586518 0.66802320
## ptratio 0.28994558 -0.39167855 0.38324756 -0.121515174 0.18893268
## black -0.38506394 0.17552032 -0.35697654 0.048788485 -0.38005064
## lstat 0.45562148 -0.41299457 0.60379972 -0.053929298 0.59087892
## medv -0.38830461 0.36044534 -0.48372516 0.175260177 -0.42732077
## rm age dis rad tax ptratio
## crim -0.21924670 0.35273425 -0.37967009 0.625505145 0.58276431 0.2899456
## zn 0.31199059 -0.56953734 0.66440822 -0.311947826 -0.31456332 -0.3916785
## indus -0.39167585 0.64477851 -0.70802699 0.595129275 0.72076018 0.3832476
## chas 0.09125123 0.08651777 -0.09917578 -0.007368241 -0.03558652 -0.1215152
## nox -0.30218819 0.73147010 -0.76923011 0.611440563 0.66802320 0.1889327
## rm 1.00000000 -0.24026493 0.20524621 -0.209846668 -0.29204783 -0.3555015
## age -0.24026493 1.00000000 -0.74788054 0.456022452 0.50645559 0.2615150
## dis 0.20524621 -0.74788054 1.00000000 -0.494587930 -0.53443158 -0.2324705
## rad -0.20984667 0.45602245 -0.49458793 1.000000000 0.91022819 0.4647412
## tax -0.29204783 0.50645559 -0.53443158 0.910228189 1.00000000 0.4608530
## ptratio -0.35550149 0.26151501 -0.23247054 0.464741179 0.46085304 1.0000000
## black 0.12806864 -0.27353398 0.29151167 -0.444412816 -0.44180801 -0.1773833
## lstat -0.61380827 0.60233853 -0.49699583 0.488676335 0.54399341 0.3740443
## medv 0.69535995 -0.37695457 0.24992873 -0.381626231 -0.46853593 -0.5077867
## black lstat medv
## crim -0.38506394 0.4556215 -0.3883046
## zn 0.17552032 -0.4129946 0.3604453
## indus -0.35697654 0.6037997 -0.4837252
## chas 0.04878848 -0.0539293 0.1752602
## nox -0.38005064 0.5908789 -0.4273208
## rm 0.12806864 -0.6138083 0.6953599
## age -0.27353398 0.6023385 -0.3769546
## dis 0.29151167 -0.4969958 0.2499287
## rad -0.44441282 0.4886763 -0.3816262
## tax -0.44180801 0.5439934 -0.4685359
## ptratio -0.17738330 0.3740443 -0.5077867
## black 1.00000000 -0.3660869 0.3334608
## lstat -0.36608690 1.0000000 -0.7376627
## medv 0.33346082 -0.7376627 1.0000000corrplot(cor_matrix, method="circle")
Correlation plot visualizes all the correlations among variables in numbers as scale.The range of correlation coefficient lies between -1 and +1. you can see the different round circles based on their strength of correlation.
cor_matrix<-cor(Boston) %>% round(digits = 2)
cor_matrix## crim zn indus chas nox rm age dis rad tax ptratio
## crim 1.00 -0.20 0.41 -0.06 0.42 -0.22 0.35 -0.38 0.63 0.58 0.29
## zn -0.20 1.00 -0.53 -0.04 -0.52 0.31 -0.57 0.66 -0.31 -0.31 -0.39
## indus 0.41 -0.53 1.00 0.06 0.76 -0.39 0.64 -0.71 0.60 0.72 0.38
## chas -0.06 -0.04 0.06 1.00 0.09 0.09 0.09 -0.10 -0.01 -0.04 -0.12
## nox 0.42 -0.52 0.76 0.09 1.00 -0.30 0.73 -0.77 0.61 0.67 0.19
## rm -0.22 0.31 -0.39 0.09 -0.30 1.00 -0.24 0.21 -0.21 -0.29 -0.36
## age 0.35 -0.57 0.64 0.09 0.73 -0.24 1.00 -0.75 0.46 0.51 0.26
## dis -0.38 0.66 -0.71 -0.10 -0.77 0.21 -0.75 1.00 -0.49 -0.53 -0.23
## rad 0.63 -0.31 0.60 -0.01 0.61 -0.21 0.46 -0.49 1.00 0.91 0.46
## tax 0.58 -0.31 0.72 -0.04 0.67 -0.29 0.51 -0.53 0.91 1.00 0.46
## ptratio 0.29 -0.39 0.38 -0.12 0.19 -0.36 0.26 -0.23 0.46 0.46 1.00
## black -0.39 0.18 -0.36 0.05 -0.38 0.13 -0.27 0.29 -0.44 -0.44 -0.18
## lstat 0.46 -0.41 0.60 -0.05 0.59 -0.61 0.60 -0.50 0.49 0.54 0.37
## medv -0.39 0.36 -0.48 0.18 -0.43 0.70 -0.38 0.25 -0.38 -0.47 -0.51
## black lstat medv
## crim -0.39 0.46 -0.39
## zn 0.18 -0.41 0.36
## indus -0.36 0.60 -0.48
## chas 0.05 -0.05 0.18
## nox -0.38 0.59 -0.43
## rm 0.13 -0.61 0.70
## age -0.27 0.60 -0.38
## dis 0.29 -0.50 0.25
## rad -0.44 0.49 -0.38
## tax -0.44 0.54 -0.47
## ptratio -0.18 0.37 -0.51
## black 1.00 -0.37 0.33
## lstat -0.37 1.00 -0.74
## medv 0.33 -0.74 1.00corrplot(cor_matrix, method="circle", type="upper", cl.pos="b", tl.pos="d", tl.cex = 0.6)
Now i will center and standardize variables and then se the summaries of the scaled variables.After that i will check the class of the boston_scaled object. Then change the object to data frame.
boston_scaled <- scale(Boston)
summary(boston_scaled)## crim zn indus chas
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563 Min. :-0.2723
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668 1st Qu.:-0.2723
## Median :-0.390280 Median :-0.48724 Median :-0.2109 Median :-0.2723
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150 3rd Qu.:-0.2723
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202 Max. : 3.6648
## nox rm age dis
## Min. :-1.4644 Min. :-3.8764 Min. :-2.3331 Min. :-1.2658
## 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366 1st Qu.:-0.8049
## Median :-0.1441 Median :-0.1084 Median : 0.3171 Median :-0.2790
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059 3rd Qu.: 0.6617
## Max. : 2.7296 Max. : 3.5515 Max. : 1.1164 Max. : 3.9566
## rad tax ptratio black
## Min. :-0.9819 Min. :-1.3127 Min. :-2.7047 Min. :-3.9033
## 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876 1st Qu.: 0.2049
## Median :-0.5225 Median :-0.4642 Median : 0.2746 Median : 0.3808
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058 3rd Qu.: 0.4332
## Max. : 1.6596 Max. : 1.7964 Max. : 1.6372 Max. : 0.4406
## lstat medv
## Min. :-1.5296 Min. :-1.9063
## 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 3.5453 Max. : 2.9865class(boston_scaled)## [1] "matrix" "array"boston_scaled<- as.data.frame(boston_scaled)Lets create a factor variable first-
Lets check the summary of the scaled crime rate and create a quantile vector of crim and print it.Then create a categorical variable ‘crime’ and look at the table of the new factor crime. After that remove the original crim from the dataset. Finally, add the new categorical value to scaled data
summary(boston_scaled$crim)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -0.419367 -0.410563 -0.390280 0.000000 0.007389 9.924110bins <- quantile(boston_scaled$crim)
bins## 0% 25% 50% 75% 100%
## -0.419366929 -0.410563278 -0.390280295 0.007389247 9.924109610crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high"))
table(crime)## crime
## low med_low med_high high
## 127 126 126 127boston_scaled <- dplyr::select(boston_scaled, -crim)
boston_scaled <- data.frame(boston_scaled, crime)Lets divide the dataset to train and test sets, so that 80% of the data belongs to the train set
n <- nrow(boston_scaled)
ind <- sample(n, size = n * 0.8)
train <- boston_scaled[ind,]
test <- boston_scaled[-ind,]
correct_classes <- test$crime
test <- dplyr::select(test, -crime)Linear Discriminant analysis Lets fit the linear discriminant analysis (LDA) on the train set.
The target variable in LDA needs to be categorical, so crime rate is the target variable and all the other variables are predictors.
LDA is based on assumptions that variables are normally distributed and each variable has the same variance. we did the scaling to the variables so this should be OK.
# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)
# print the lda.fit object
lda.fit## Call:
## lda(crime ~ ., data = train)
##
## Prior probabilities of groups:
## low med_low med_high high
## 0.2549505 0.2475248 0.2673267 0.2301980
##
## Group means:
## zn indus chas nox rm age
## low 1.04343196 -0.8822584 -0.08120770 -0.8963915 0.3982842 -0.9203203
## med_low -0.07776353 -0.2990194 0.04263895 -0.5761771 -0.1061665 -0.3652688
## med_high -0.37448920 0.2113431 0.23803578 0.4039967 0.1617825 0.4232494
## high -0.48724019 1.0173143 -0.06065701 1.0104611 -0.4319335 0.8123991
## dis rad tax ptratio black lstat
## low 0.9219518 -0.6797017 -0.7101905 -0.44988670 0.38273133 -0.76461330
## med_low 0.3368015 -0.5500476 -0.4865228 -0.08477516 0.31429335 -0.14977651
## med_high -0.3984171 -0.4076377 -0.3009895 -0.37250800 0.08103825 -0.05396649
## high -0.8686609 1.6361396 1.5126335 0.77846144 -0.77989700 0.81304005
## medv
## low 0.47273317
## med_low 0.04503523
## med_high 0.23724952
## high -0.64367034
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3
## zn 0.07713607 0.669931724 -0.93155652
## indus 0.04209589 -0.164886404 0.15234811
## chas -0.08682276 -0.012943618 0.09626586
## nox 0.34246364 -0.821808186 -1.27271655
## rm -0.10863073 -0.129260809 -0.16756710
## age 0.22622361 -0.330535144 -0.21506781
## dis -0.08235521 -0.245442165 0.06111034
## rad 3.13566288 0.921028504 -0.22879119
## tax 0.02757234 0.009131622 0.60584712
## ptratio 0.13077342 -0.025325718 -0.17176686
## black -0.13581331 0.022095261 0.07061716
## lstat 0.24708262 -0.221805806 0.58905552
## medv 0.24358364 -0.459633802 0.02212599
##
## Proportion of trace:
## LD1 LD2 LD3
## 0.9401 0.0465 0.0133# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(train$crime)
# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)
Proportion of trace is the variance of between groups , here LD1 94% explains the between groups variance.
The arrows are drawn based on the coefficients. You can find 5 distinct classess.
Predict LDA
Lets predict the classes with the LDA model -test data.
# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)
# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)## predicted
## correct low med_low med_high high
## low 11 11 2 0
## med_low 2 20 4 0
## med_high 0 7 10 1
## high 0 0 0 34the model can not predict very well
distance measures
# load MASS and Boston
library(MASS)
data('Boston')
# euclidean distance matrix
dist_eu <- dist(Boston)
# look at the summary of the distances
summary(dist_eu)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.119 85.624 170.539 226.315 371.950 626.047# manhattan distance matrix
dist_man <- dist(Boston, method = 'manhattan')
# look at the summary of the distances
summary(dist_man)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 2.016 149.145 279.505 342.899 509.707 1198.265Distances are totally different between these two distance methods.
K-means clustering “It is an unsupervised method, that assigns observations to groups or clusters based on similarity of the objects.” In clustering, you don’t know the number of classes beforehand. “K-means calculates distances between centroids and datapoi
# Boston dataset is available
# k-means clustering
km <-kmeans(Boston, centers = 4)
# plot the Boston dataset with clusters
pairs(Boston, col = km$cluster)
determine the k
Lets find the best number of clusters:
# Boston dataset is available
set.seed(123)
# determine the number of clusters
k_max <- 10
# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(Boston, k)$tot.withinss})
# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')
# k-means clustering
km <-kmeans(Boston, centers = 2)
# plot the Boston dataset with clusters
pairs(Boston, col = km$cluster)
Optimal number of clusters might be 2, because there the total within cluster sum of squares (WCSS) changes radically.
model_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)Lets make a 3D plot
library(plotly)##
## Attaching package: 'plotly'## The following objects are masked from 'package:plyr':
##
## arrange, mutate, rename, summarise## The following object is masked from 'package:MASS':
##
## select## The following object is masked from 'package:ggplot2':
##
## last_plot## The following object is masked from 'package:stats':
##
## filter## The following object is masked from 'package:graphics':
##
## layoutplot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers')## Warning: `arrange_()` is deprecated as of dplyr 0.7.0.
## Please use `arrange()` instead.
## See vignette('programming') for more help
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_warnings()` to see where this warning was generated.(more chapters to be added similarly as we proceed with the course!)